Question: Graph this system of equations and solve. $-x-3y = 9$ $-6x-3y = -6$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: Convert the first equation, $-x-3y = 9$ , to slope-intercept form. $y = -\dfrac{1}{3} x - 3$ The y-intercept for the first equation is $-3$ , so the first line must pass through the point $(0, -3)$ The slope for the first equation is $-\dfrac{1}{3}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $3$ positions to the right. $3$ positions to the right. Graph the blue line so it passes through $(0, -3)$ and $(3, -4)$ Convert the second equation, $-6x-3y = -6$ , to slope-intercept form. $y = -2 x + 2$ The y-intercept for the second equation is $2$ , so the second line must pass through the point $(0, 2)$ The slope for the second equation is $-2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) You must also move $1$ positions to the right. $1$ position to the right. $2$ positions down from $(0, 2)$ is $(1, 0)$ Graph the green line so it passes through $(0, 2)$ and $(1, 0)$ The solution is the point where the two lines intersect. The lines intersect at $(3, -4)$.